Math 120A HW 6 Solutions
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Let $\sigma$ and $\tau$ be disjoint permutations of a set $A$.
Consider an arbitrary element $x \in A$.
- Case 1: $x \in \text{supp}(\sigma)$,
meaning that $\sigma(x) \ne x$. Applying $\sigma$ to both sides and using injectivity, we get $\sigma(\sigma(x)) \ne \sigma(x)$,
so $\sigma(x) \in \text{supp}(\sigma)$ also. (This is a general argument showing that the support of any permutation is closed under that permutation.)
Because the supports of $\sigma$ and $\tau$ are disjoint, this means that $x, \sigma(x) \notin \text{supp}(\tau)$.
Therefore we have
$\sigma(\tau(x)) = \sigma(x)$ and also
$\tau(\sigma(x)) = \sigma(x)$, so $\sigma\tau$ and $\tau\sigma$ agree in this case.
- Case 2: $x \in \text{supp}(\tau)$. By a similar argument to case 1, we have $\sigma(\tau(x)) = \tau(x) = \tau(\sigma(x))$, so $\sigma\tau$ and $\tau\sigma$ agree in this case also.
- Case 3: $x \notin \text{supp}(\sigma)$ and $x \notin
\text{supp}(\tau)$. In this case $\sigma(\tau(x))$ and $\tau(\sigma(x))$
are both equal to $x$, so again $\sigma\tau$
and $\tau\sigma$ agree.
Therefore $\sigma\tau = \tau \sigma$.
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To show that $D_n$ is closed under multiplication (composition) there are four cases to consider.
Let's redefine $i \bmod n$ in this problem to mean the unique number in
the set $\{1,\ldots,n\}$ that differs from $i$ by a multiple of $n$,
rather than the unique number in the set $\{0,\ldots,n-1\}$ that differs
from $i$ by a multiple of $n$. (This doesn't affect the logical structure
of the proof at all, but it simplifies the notation.)
- Case 1: The product of two rotations $\sigma_j$ and $\sigma_k$.
For all $i \in \{1,\ldots,n\}$ we have $$(\sigma_j\sigma_k)(i) =
\sigma_j(\sigma_k(i)) = j + (k + i \bmod n) \bmod n = (j+k \bmod n) +
i \bmod n = \sigma_{j+k \bmod n}(i),$$ so $\sigma_j\sigma_k = \sigma_{j+k
\bmod n} \in D_n$.
- Case 2: The product of a rotation $\sigma_j$ and a reflection $\rho_k$.
For all $i \in \{1,\ldots,n\}$ we have $$(\sigma_j\rho_k)(i) =
\sigma_j(\rho_k(i)) = j + (k - i \bmod n) \bmod n = (j+k \bmod n) -
i \bmod n = \rho_{j+k \bmod n}(i),$$ so $\sigma_j\rho_k = \rho_{j+k
\bmod n} \in D_n$.
- Case 3: The product of a reflection $\rho_k$ and a rotation $\sigma_j$ (in the other order from case 2.)
For all $i \in \{1,\ldots,n\}$ we have $$(\rho_k\sigma_j)(i) =
\rho_k(\sigma_j(i)) = k - (j + i \bmod n) \bmod n = (k-j \bmod n) - i
\bmod n = \rho_{k-j \bmod n}(i),$$ so $\rho_k\sigma_j = \rho_{k-j
\bmod n} \in D_n$.
- Case 4: The product of two reflections $\rho_j$ and $\rho_k$.
For all $i \in \{1,\ldots,n\}$ we have $$(\rho_j\rho_k)(i) =
\rho_j(\rho_k(i)) = j - (k - i \bmod n) \bmod n = (j-k \bmod n) +
i \bmod n = \sigma_{j-k \bmod n}(i),$$ so $\rho_j\rho_k = \sigma_{j-k
\bmod n} \in D_n$.
- I will write the elements in cycle notation, but the two-row notation is fine too.
- The orbits of the identity element are $\{1\},\{2\},\{3\}$.
- The orbits of $(1\;2)$ are $\{1,2\},\{3\}$.
- The orbits of $(1\;3)$ are $\{1,3\},\{2\}$.
- The orbits of $(2\;3)$ are $\{2,3\},\{1\}$.
- The only orbit of $(1\;2\;3)$ is $\{1,2,3\}$.
- The only orbit of $(1\;3\;2)$ is also $\{1,2,3\}$.
- Again I will write the elements in cycle notation, but the two-row notation is fine too.
- $(1\;2)(3\;4)$, $(1\;3)(2\;4)$, $(1\;4)(2\;3)$.
- $(2\;3\;4)$, $(2\;4\;3)$, $(1\;3\;4)$, $(1\;4\;3)$, $(1\;2\;4)$, $(1\;4\;2)$, $(1\;2\;3)$, $(1\;3\;2)$.
- In $S_6$, consider the following product of cycles: $\sigma = (2\;6\;1)(3\;6)(1\;2\;4)$.
- $\sigma =
\begin{pmatrix}
1&2&3&4&5&6\\
6&4&1&2&5&3
\end{pmatrix}$.
- $\sigma = (1\;6\;3)(2\;4)$.
- Write each element of the dihedral group $D_6$ as a product of disjoint cycles. (For example, $\sigma_3 = (1\;4)(2\;5)(3\;6)$.)
- $\sigma_1 = (1\;2\;3\;4\;5\;6)$
- $\sigma_2 = (1\;3\;5)(2\;4\;6)$
- $\sigma_3 = (1\;4)(2\;5)(3\;6)$
- $\sigma_4 = (1\;5\;3)(2\;6\;4)$
- $\sigma_5 = (1\;6\;5\;4\;3\;2)$
- $\sigma_6 = \text{identity}$ (empty product)
- $\rho_1 = (1\;6)(2\;5)(3\;4)$
- $\rho_2 = (2\;6)(3\;5)$
- $\rho_3 = (1\;2)(3\;6)(4\;5)$
- $\rho_4 = (1\;3)(4\;6)$
- $\rho_5 = (1\;4)(2\;3)(5\;6)$
- $\rho_6 = (1\;5)(2\;4)$